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Does the coriolis force has any measurable effect in free fall from large heights?

Take for example the sky diving experiment by F. Baumgartner who started from a height of about 40 km above New Mexico. Does the coriolis force has any measurable effect in this experiment?

I think there should be a deviation to the east and another one to the south (which should be smaller in size). But I don't have an idea how large the deviation would be.

How can one calculate the size of the deviations?

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forgive my ignorance but isn't the coriolis force only a matter of changing the frame of reference? The coriolis force shouldn't er... have a physical effect on Baumgartner right? – drN Oct 21 '12 at 23:30
@drN, the coriolis force won't have an effect on Baumgartner in an inertial frame. But it will cause him to hit the ground at a different point than directly below where he began his fall. – Colin McFaul Oct 22 '12 at 1:10
@ColinMcFaul True that! – drN Oct 22 '12 at 12:46

Yes it does. If you fix a (non inertial) frame of reference whose origin is on Earth's surface, at latitude $\lambda$, then a freely falling particle of mass $m$ has the equation of motion, $$m\frac{d^2\vec r}{dt^2}=m\vec g_{ef}-2m\vec\omega\times\vec r,$$ where $$\vec g_{ef}=\vec g-\vec\omega\times\left[\omega\times(\vec R+\vec r)\right]\approx \vec g-\vec\omega\times(\omega\times\vec r),$$ is the effective gravity which takes into account the centrifugal force and $\vec R$ is the vector from the center of the Earth to the origin of the non inertial frame. The vector $\vec omega$ is the angular velocity of the Earth. Solving for $\vec r$ one gets $$\vec r(t)=\vec v(0)t+\frac{\vec g_{ef}t^2}{2}-2\omega\times\int_0^t \vec r(t)dt.$$ This equation can be soved order by order, the first order being $$\vec r(t)=\vec v(0)t+\frac{\vec g_{ef}t^2}{2}-2\omega\times\left(\frac{\vec v(0)t^2}{2}+\frac{\vec g_{ef}t^3}{6}\right).$$ If the particle has $\vec v(0)=\vec 0$, the deflection relative to the effective vertical (defined by the direction of $\vec g_{ef}$) is given by $$\Delta x=\frac{\vec\omega\times\vec g_{ef}t^3}{3}=\frac{g_{ef}\omega t^3\cos\lambda}{3}\hat i,$$ where $\hat i$ is directed east (tangentially to the circle of latitude). Plugging in the approximate time to fall from the height $h$, $t=\sqrt{2h/g_{ef}}$, $$\Delta x=\frac{1}{3}\left(\frac{8h}{g_{ef}}\right)^{1/2}\omega\cos\lambda\approx \frac{1}{3}\left(\frac{8h^3}{g}\right)^{1/2}\omega\cos\lambda,$$ since $g_{ef}$ already is $g_{ef}=g+\mathcal O(\omega^2)$. The angular velocity of the Earth is $\omega\approx 7.3\cdot 10^{-5}\, \mathrm{s}^{-1}$, so a free fall of order $10^4\, \mathrm m$ would give a deviation of order $10^1\, \mathrm m$. Notice however that this not take the drag force into account.

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